moldovan wrote: Find all functions $ f: \mathbb{R}^{ + }\rightarrow \mathbb{R}$ that satisfy: $ f(x)f(y) = f(xy) + 2005 \left( \frac {1}{x} + \frac {1}{y} + 2004 \right)$ for all $ x,y > 0.$ $ y=1$ $ \implies$ $ f(x)(f(1)-1)=2005(2005+\frac 1x)$. So $ f(1)\neq 1$ and $ f(x)=a(2005+\frac 1x)$ Plugging back in the original equation, we get $ a=1$ and the unique solution $ \boxed{f(x)=2005+\frac 1x}$